The emission of two successive beta particles from the nucleus \(^{32} _{15} P\) will produce
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Correct Answer: Option C
Explanation:
\(^{32} _{15}\)P \(\to\) 2\(^{0} _{-1}\)e + \(^{a} _{b}\)X
32 = 2(0) + a
a = 32
15 = 2(-1) + b
b = 15 + 2
b = 17
\(\therefore\) \(^{a} _{b}\)X = \(^{32} _{17}\)Cl
\(^{32} _{15}\)P \(\to\) 2\(^{0} _{-1}\)e + \(^{a} _{b}\)X
32 = 2(0) + a
a = 32
15 = 2(-1) + b
b = 15 + 2
b = 17
\(\therefore\) \(^{a} _{b}\)X = \(^{32} _{17}\)Cl